Question

If $x+y-z=4$ and $x^2+y^2+z^2=50$, find the value of $xy-yz-zx$

• A. $-13$
• B. $-17$
• C. $18$
• D. $21$

Answer 1

The correct option is B $-17$

Given,
$x+y+z=4$ and $x^2+y^2+z^2=50$
We know,
$(a+b+c{)}^2=(a^2+b^2+c^2)+2(ab+bc+ca)$
Here,
$a=x,b=y,c=-z$
Thus,
$(x+y-z{)}^2=x^2+y^2+z^2+2(xy-yz-xz)$
$=>4^2=50+2(xy-yz-xz)$
$=>\frac{16-50}{2}=(xy-yz-xz)$
$=>(xy-yz-xz)=-17$